A128535 -id:A128535 - cp's OEIS frontend

A114695 Three consecutive elements of the sequence built from a quadratic form over four consecutive Fibonacci numbers A000045.

2, 2, 4, 104, 143, 169, 4895, 6764, 7921, 229970, 317810, 372100, 10803704, 14930351, 17480761, 507544127, 701408732, 821223649, 23843770274, 32951280098, 38580030724, 1120149658760, 1548008755919, 1812440220361

Offset: 0

Roger L. Bagula, Feb 21 2006

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (0,0,48,0,0,-48,0,0,1).

Cf. A000032, A000045.

F[n_]:= Fibonacci[n]; Flatten[Table[{F[4*n+2]*F[4*n+3], (F[4*n]+F[4*n+2])*F[4*n+ 3], F[4*n+3]^2}, {n, 0, 12}]] (* modified by G. C. Greubel, May 24 2021 *)
With[{m = Floor[n/3], F = Fibonacci}, Table[F[4*m+3]*(4*F[4*m+2] -(Mod[n^2,3]*F[4*m +2] +Mod[(n+2)^2, 3]*LucasL[4*m+1] +Mod[(n+1)^2, 3]*F[4*m+3])), {n, 0, 40}]] (* G. C. Greubel, May 24 2021 *)

f=fibonacci;
def A114695(n): return f(4*(n//3)+3)*( 4*f(4*(n//3)+2) - ((n^2%3)*f(4*(n//3)+2) + ((n+2)^2%3)*(f(4*(n//3)+2) + f(4*(n//3))) + ((n+1)^2%3)*f(4*(n//3)+3) ) )
[A114695(n) for n in (0..40)] # G. C. Greubel, May 24 2021

a(3*n) = Fibonacci(4*n+2)*Fibonacci(4*n+3).
a(3*n+1) = Lucas(4*n+1)*Fibonacci(4*n+3).
a(3*n+2) = Fibonacci(4*n+3)*Fibonacci(4*n+3).
From R. J. Mathar, Apr 16 2009: (Start)
a(3*n) = A001654(4*n+2).
a(3*n+1) = A128535(4*n+3).
a(3*n+2) = A007598(4*n+3).
G.f.: (2+2*x+4*x^2+8*x^3+47*x^4-23*x^5-x^6-4*x^7+x^8)/((1-x)*(1+x+x^2)*(1-47*x^3+x^6)).
a(n) = 48*a(n-3) - 48*a(n-6) + a(n-9). (End)
a(n) = F(4*m+3)*( 4*F(4*m+2) - ((n^2 mod 3)*F(4*m+2) + ((n+2)^2 mod 3)*Lucas(4*m+1) + ((n+1)^2 mod 3)*F(4*m+3)) ), where m = floor(n/3) and F = Fibonacci. - G. C. Greubel, May 24 2021

Edited by the Associate Editors of the OEIS, Sep 02 2009

A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

Original entry on oeis.org

1, 1, -2, 2, -2, 5, -9, 13, -20, 34, -56, 89, -143, 233, -378, 610, -986, 1597, -2585, 4181, -6764, 10946, -17712, 28657, -46367, 75025, -121394, 196418, -317810, 514229, -832041, 1346269, -2178308, 3524578, -5702888

Offset: 0

Creighton Dement, Feb 23 2006

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,0,-1,1).

Cf. A000045, A001519, A006498, A056594, A115008, A116698, A116699, A128535.

Cf. A186679 (first differences).

a116697 n = a116697_list !! n
a116697_list = [1,1,-2,2]
               ++ (zipWith (-) a116697_list
                               $ zipWith (+) (tail a116697_list)
                                             (drop 3 a116697_list))
a128535_list = 0 : (map negate $ map a116697 [0,2..])
a001519_list = 1 : map a116697 [1,3..]
a186679_list = zipWith (-) (tail a116697_list) a116697_list
a128533_list = map a186679 [0,2..]
a081714_list = 0 : (map negate $ map a186679 [1,3..])
a075193_list = 1 : -3 : (zipWith (+) a186679_list $ drop 2 a186679_list)
-- Reinhard Zumkeller, Feb 25 2011

A116697:= func< n | (-1)^Floor((n+1)/2)*(1+(-1)^n)/2 -(-1)^n*Fibonacci(n) >;
[A116697(n): n in [0..50]]; // G. C. Greubel, Jun 08 2025

LinearRecurrence[{-1,0,-1,1},{1,1,-2,2},40] (* Harvey P. Dale, Nov 04 2011 *)

def A116697(n): return (-1)^(n//2)*((n+1)%2) - (-1)^n*fibonacci(n)
print([A116697(n) for n in range(51)]) # G. C. Greubel, Jun 08 2025

G.f.: (1 + 2*x - x^2 + x^3)/((1 + x^2)*(1 + x - x^2)).
a(2*n+1) = A000045(2*n+1) = A001519(n).
a(2*n) = - A128535(n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A056594(n) - (-1)^n*A000045(n). - Bruno Berselli, Feb 26 2011
E.g.f.: cos(x) + (2/sqrt(5))*exp(-x/2)*sinh(sqrt(5)*x/2). - G. C. Greubel, Jun 08 2025

A128533 a(n) = F(n)*L(n+2) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

0, 4, 7, 22, 54, 145, 376, 988, 2583, 6766, 17710, 46369, 121392, 317812, 832039, 2178310, 5702886, 14930353, 39088168, 102334156, 267914295, 701408734, 1836311902, 4807526977, 12586269024, 32951280100, 86267571271, 225851433718, 591286729878

Offset: 0

Axel Harvey, Mar 08 2007

Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1): if k = 0 then sequence is A001906, if k = 1 it is A081714.

			a(4) = 54 because F(4)*L(6) = 3*18.
G.f. = 4*x + 7*x^2 + 22*x^3 + 54*x^4 + 145*x^5 + 376*x^6 + 988*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001906, A081714, A128534, A128535.

List([0..30], n -> Fibonacci(2*(n+1)) + (-1)^(n+1)); # G. C. Greubel, Jan 07 2019

[Fibonacci(n)*Lucas(n+2): n in [0..30]]; // Vincenzo Librandi, Feb 20 2013

with(combinat); A128533:=n->fibonacci(2*n+2)+(-1)^(n+1); seq(A128533(k),k=0..50); # Wesley Ivan Hurt, Oct 19 2013

LinearRecurrence[{2,2,-1}, {0,4,7}, 40] (* Vincenzo Librandi, Feb 20 2013 *)
a[n_]:= Fibonacci[2n+2] -(-1)^n; (* Michael Somos, May 26 2014 *)

vector(30, n, n--; fibonacci(2*(n+1)) + (-1)^(n+1)) \\ G. C. Greubel, Jan 07 2019

[fibonacci(2*(n+1)) + (-1)^(n+1) for n in (0..30)] # G. C. Greubel, Jan 07 2019

a(n) = F(2*(n+1)) + (-1)^(n+1), assuming F(0) = 0 and L(0) = 2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(4-x)/((1+x)*(x^2-3*x+1)). (End)
a(n) = A186679(2*n). - Reinhard Zumkeller, Feb 25 2011
a(-n) = - A128535(n). - Michael Somos, May 26 2014
0 = a(n)*(+4*a(n) + a(n+1) - 17*a(n+2)) + a(n+1)*(-14*a(n+1) + a(n+2)) + a(n+2)*(+4*a(n+2)) for all n in Z. - Michael Somos, May 26 2014

More terms from Vincenzo Librandi, Feb 20 2013

A128534 a(n) = Fibonacci(n)*Lucas(n-1).

Original entry on oeis.org

0, 2, 1, 6, 12, 35, 88, 234, 609, 1598, 4180, 10947, 28656, 75026, 196417, 514230, 1346268, 3524579, 9227464, 24157818, 63245985, 165580142, 433494436, 1134903171, 2971215072, 7778742050, 20365011073, 53316291174, 139583862444, 365435296163, 956722026040, 2504730781962

Offset: 0

Axel Harvey, Mar 08 2007

Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1). If k=0 the sequence is A001906; if k=1 it is A081714.
a(n) is the maximum area of a quadrilateral with lengths of sides in order F(n), F(n), L(n-1), L(n-1) for n>1. - J. M. Bergot, Jan 28 2016
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Lucas numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			a(5) = 35 because F(5)*L(4) = 5*7.

Robert Israel, Table of n, a(n) for n = 0..2370
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A001906, A081714, A128533, A128535.

[Fibonacci(n)*Lucas(n-1): n in [0..30]]; // G. C. Greubel, Dec 21 2017

seq(combinat:-fibonacci(2*n-1)+(-1)^(n+1),n=0..50); # Robert Israel, Jan 28 2016

Table[Fibonacci[n] LucasL[n - 1], {n, 0, 31}] (* Michael De Vlieger, Jan 29 2016 *)

concat( 0, Vec(-x*(-2+3*x)/((1+x)*(x^2-3*x+1)) + O(x^40))) \\ Michel Marcus, Jan 28 2016

a(n) = F(2*n - 1) + (-1)^(n+1), assuming F(0)=0 and L(0)=2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(2-3*x)/((1+x)*(x^2-3*x+1)). (End)
a(n) = (2^(-1-n)*(-5*(-1)^n*2^(1+n) - (-5+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Apr 05 2016
a(n+1) = A081714(n) + 2*(-1)^n. - A.H.M. Smeets, Feb 26 2022

More terms from Michel Marcus, Jan 28 2016

A236144 a(n) = F(floor( (n+3)/2 )) * L(floor( (n+2)/2 )) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

2, 2, 1, 2, 6, 9, 12, 20, 35, 56, 88, 143, 234, 378, 609, 986, 1598, 2585, 4180, 6764, 10947, 17712, 28656, 46367, 75026, 121394, 196417, 317810, 514230, 832041, 1346268, 2178308, 3524579, 5702888, 9227464, 14930351, 24157818, 39088170, 63245985, 102334154

Offset: 0

Michael Somos, Jan 19 2014

			G.f. = 2 + 2*x + x^2 + 2*x^3 + 6*x^4 + 9*x^5 + 12*x^6 + 20*x^7 + 35*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..2500
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000032, A000045, A115008, A115339, A128534, A128535.

m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(2-x^2-x^3)/(1-x-x^3-x^4)); // G. C. Greubel, Aug 07 2018

a[ n_] := Fibonacci[ Quotient[ n + 3, 2]] LucasL[ Quotient[ n, 2]];
CoefficientList[Series[(2-x^2-x^3)/(1-x-x^3-x^4), {x, 0, 60}], x] (* G. C. Greubel, Aug 07 2018 *)

{a(n) = fibonacci( (n+3)\2 ) * (fibonacci( n\2+1 ) + fibonacci( n\2-1 ))};

x='x+O('x^60); Vec((2-x^2-x^3)/(1-x-x^3-x^4)) \\ G. C. Greubel, Aug 07 2018

G.f.: (2 - x^2 - x^3) / (1 - x - x^3 - x^4) = (1 - x) * (2 + 2*x + x^2) / ((1 + x^2) * (1 - x - x^2)).
a(n) = a(n-1) + a(n-3) + a(n-4) for all n in Z.
0 = a(n)*a(n+2) + a(n+1)*(+a(n+2) -a(n+3)) for all n in Z.
a(n) = A115008(n+2) - A115008(n+1).
a(n) = A115339(n) * A115339(n-1).
a(2*n - 1) = F(n+1) * L(n-1) = A128535(n+1). a(2*n) = F(n+1) * L(n) = A128534(n+1).
a(n) = A000045(n+1)+A057077(n). - R. J. Mathar, Sep 24 2021

cp's OEIS Frontend

A114695 Three consecutive elements of the sequence built from a quadratic form over four consecutive Fibonacci numbers A000045.

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A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

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A128533 a(n) = F(n)*L(n+2) where F=Fibonacci and L=Lucas numbers.

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A128534 a(n) = Fibonacci(n)*Lucas(n-1).

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A236144 a(n) = F(floor( (n+3)/2 )) * L(floor( (n+2)/2 )) where F=Fibonacci and L=Lucas numbers.

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